Conference rusure::math

     In the first case, add the integral over z=-1 to z=1:

          Int(top arc) + Int(z=-1 to 1) = Int(arc + z=-1 to 1)

     The r.h.s. integral is over a closed contour that does not
     include the only singularity of 1/(z-2), which is at z=2.
     Therefore the integral over the closed contour is zero, and
     so

          Int(top arc) + Int(z=-1 to z=1 of dz/(z-2)) = 0

     and this integral is easy to solve.

     In the second case, add the integral over the bottom half of
     the circle:

          Int(top arc) + Int(bottom arc) = Int(circle)

     The right hand side integral is over a closed contour that
     does include the only singularity of 1/(z-2).  Therefore the
     integral over the closed contour is whatever it is supposed
     to be [I took this course in the fall of 1977].  I think it
     is 2 * pi * i times whatever it is called, which here is 1.
     Now you have to use some kind of symmetry argument to
     determine how Int(bottom arc) relates to Int(top arc), and I
     would be surprised if they were other than equal or
     negatives of each other.  Plug the result into

          Int(top arc) + (+/- 1) * Int(top arc) = 2 pi i

     Dan

    Thanks for the reply Dan.  About two hours after writing the note
    I thought I stumbled on the answer...though I'm not sure how it
    tracks with your approach.
    
    I took the integral of dz/(z-2)=ln(z-2) and I then substituted
    z=r*(cos(theta)+i*sin(theta)).  Where r=1 in case a and r=3 in case
    b.  Theta goes from 0 to pi as z goes from 1 to -1 (or 3 to -3 in
    case b).
    
    The first subtle point that eluded me (unless I'm still all wrong)
    is that the argument of the natural log can be negative in the complex
    domain. In fact Ln(z)=ln|z| + i*Arg(z), where the capital lettered
    terms refer to the principal value of the complex natural log.
    
    The second point that eluded me, for case b is that the semicircular
    arc in the upper half plane from z=3 to z=-3 NEVER crosses the
    singularity at z=2 (very counter intuitive to me for weeks).  The
    singularity would be at z=2 which is purely real and the only purely
    real values along the given contour are 3 and -3.
    
    Well, if my midnight reasoning and my math are right the answer to
    case a is 1.099 + i*pi and for case b is 1.61 + i*pi.
    
    regards...Ira
         

    I think the first integral should be log(3); since the semicircular
    arc can be deformed onto the real axis without encountering the
    pole at z = 2, the integral drops out as a simple real integral.

	INT(dz/(z-2))[1,-1] = log(-3)-log(-1) = log(3) ~= 1.09861

    The second integral is probably best done by deforming the path
    out along the real axis towards infinity, with a large semicircular
    arc connecting the ends.  The integral of the large arc tends to
    zero, while the pieces on the real axis can be done as a pair of
    real integrals, and we should get log(-5), about 1.60944+j*pi.

    When all else fails you can always fall back on the definition of
    a contour integral as a line integral along a parameterized arc
    lying in the plane;  in particular, the substitution z = exp(j*t)
    and dz = j*exp(j*t) dt could bring these integrals into real form...

    - Jim

    In reading .1, a pitfall to avoid is that symmetric paths may get
    different cancellation of real and / or imaginary parts depending on the
    orientation of the paths.  For example a contour in the lower half
    plane will be along a path which is the complex conjugate of the
    upper path, and simple multiplication by -1 doesn't work in combining
    the pieces...

    - Jim

    I should clarify two points in my reply since some muddle headed
    reasoning was involved, though the answers are correct.

    Reparameterize both integrals via the substitution z -> 2-z.
    Now both contours are reflectd and displaced and appear in the
    lower half plane, while the pole is at the origin.

    The first integral can be deformed onto the real line and
    is a simple real integral from 1 to 3, or log(3)-log(1) = log(3).

    The second contour may be deformed to a semicircular arc of radius
    1 from -1 to 1, which yields j*pi, and a real integral from 1 to 5,
    which yields log(5) as above.

    - Jim

>> .1     Now you have to use some kind of symmetry argument to
>>        determine how Int(bottom arc) relates to Int(top arc), and I
>>        would be surprised if they were other than equal or
>>        negatives of each other.

>> .4     In reading .1, a pitfall to avoid is that symmetric paths may get
>>        different cancellation of real and / or imaginary parts depending on
>>        the orientation of the paths.

     Surprise!  If you substitute z=re^it (using t for theta;
     here r = 3) then the integral over the top arc is

          Int(t = 0 to pi of (rie^it dt)/(re^it - 2))

     and over the bottom arc is

          Int(t = pi to 2 pi of (rie^it dt)/(re^it - 2))

     Substitute t-pi for t and note that e^(i(t-pi)) = (e^it)/(e^(i pi))
     = -e^it, and the integral over the bottom arc becomes

          Int(t = 0 to pi of (rie^it dt)/(re^it + 2))

     Note that the "- 2" has become a "+ 2" for this integral.
     So the integrals over the two arcs do not combine as nicely
     as I had expected.

>> .3    The second integral is probably best done by deforming the path
>>       out along the real axis towards infinity, with a large semicircular
>>       arc connecting the ends.  The integral of the large arc tends to
>>       zero,

     As a general technique, closing the contour "at infinity"
     should be remembered.  But it doesn't apply here -- the
     integral over an arc of radius r as r -> oo does not vanish,
     because (rie^it dt)/(re^it - 2) has a factor of r in both
     the numerator and the denominator.  However, if you complete
     the contour in the second case by going along the real axis
     from -3 to 2-epsilon, clockwise over the top half arc of
     radius epsilon to 2+epsilon, and finally along the real axis
     from 2+epsilon to 3, then you can solve this.

     My final results, the same as Jim's in .5:

          Int(top arc of radius 1) = ln 3
          Int(top arc of radius 3) = ln 5

     What are the answers if the arc has radius r, for 0 < r < 2
     and for 2 < r < oo ?

     Dan

    The mistake pointed out in .6 is why I entered a clarification.

    Anyhow, substituting z -> 2-z simplifies things, without any
    overly 'complex' limiting arguments.  Simply do a real integral
    for r < 2, and a semicircular arc of radius r-2 in the lower plane
    which yields a constant j*pi, plus a real integral for the real part,
    for r > 2.

    So in the general case we should have

	0 <= r < 2		ln(2+r) - ln(2-r)
	2 <  r < oo		ln(2+r) - ln(2-r) + j*pi

    For r = 1 we get ln(3) - ln(1) = ln(3)
    For r = 3 we get ln(5) - ln(1) + j*pi

    We could even arm wave about crossing a branch line of the log
    since when r > 2, ln(2-r) should get a j*pi added in!

    The integral has a logarithmic singularity if r = 2.

    - Jim

    Re .-1:
    
>>    For r = 3 we get ln(5) - ln(1) + j*pi
    
    I did not get a j*pi term for r=3.  The result was ln(5).
    I don't recall crossing any branch lines, either.
    
    When integrating dz/(z-2) for z > 2, there is no problem with the
    logarithms.  When integrating dz/(z-2) for z < 2, I multiplied
    both "sides" by minus one to get the integral of -dz/(2-z), and
    again there were no problems with the logarithms.  There is no
    "branch" of 1/(z-2) to cause any problems.
    
    Dan

    Re last few:
    
    Oops.  For the r > 2 case I forgot to add in the integral over
    the clockwise half circle of radius epsilon around z=2.  So my
    [final!] results are also
    
         0 <= r < 2       ln(2+r) - ln(2-r)
         2 < r < oo       ln(r+2) - ln(r-2) + i pi
    
    Dan

    I'm sorry about poor use of terminology - branch line was not quite
    what was intended.

    When you integrate a differential along a contour there results by
    analytic continuation a path issuing from the origin.  This always
    works, whether the differential is multiple valued, or if branches
    occur due to the integration - but you do have to specify the path.

    In the first case, we integrate dz/(z-2) along a path to the left
    of the pole and the running integral has an imaginary part equal
    to the counterclockwise change in angle of a radius vector from the
    pole to the moving point on the countour.  At the end of the contour,
    that net angle has returned to zero, so the integral is real.

    But in the second case the radius angle made a net change of pi
    radians as we followed the contour, this is how the value of pi
    magically appeared in the result.

    Harry Nyquist had a keen insight into the behaviour of the complex
    plane...

    - Jim

    A view of what is going on in the solution to .0:
    
o       If you integrate the complex function
o    
o                                     n
o                             f(z) = z      n = ..., -2, -1, 0, 1, 2, ...
o    
o       around a "nice" closed contour, the result is zero UNLESS ...
o       ... unless n = -1 and the "pole" at z=0 is inside the closed
o       contour.  In that case the result is 2 * pi * sqrt(-1)
    
    So to compute the contour integral over a path that isn't closed,
    you find a path that "closes" the given one, and such that you can
    evaluate the integral over it [by whatever means].  Then you figure
    out from the above what the integral over the closed path is.  Finally
    you subtract out the part that came from your auxiliary path.
    
    All the rest is just filling in the details. (-:
    
    There must be lots of books on the subject; you just have to review
    one before the test, and actually try some of the example problems,
    then it will all come back to you.
    
    Dan

    RE. -.1 Great pun.
    
    "I think he's got a pole in the right half plane." - an old insult
    I used to use about my Linear Systems professor.
    
    Martin H.